3.2.20 \(\int \frac {\tan ^2(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx\) [120]

Optimal. Leaf size=126 \[ -\frac {3 \sec (e+f x)}{5 f \sqrt [3]{a+a \sin (e+f x)}}+\frac {11 \sqrt [6]{2} \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{15 f \sqrt [6]{1+\sin (e+f x)} \sqrt [3]{a+a \sin (e+f x)}}+\frac {4 \sec (e+f x) (a+a \sin (e+f x))^{2/3}}{5 a f} \]

[Out]

-3/5*sec(f*x+e)/f/(a+a*sin(f*x+e))^(1/3)+11/15*2^(1/6)*cos(f*x+e)*hypergeom([1/2, 5/6],[3/2],1/2-1/2*sin(f*x+e
))/f/(1+sin(f*x+e))^(1/6)/(a+a*sin(f*x+e))^(1/3)+4/5*sec(f*x+e)*(a+a*sin(f*x+e))^(2/3)/a/f

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Rubi [A]
time = 0.15, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2791, 2934, 2731, 2730} \begin {gather*} \frac {11 \sqrt [6]{2} \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{15 f \sqrt [6]{\sin (e+f x)+1} \sqrt [3]{a \sin (e+f x)+a}}+\frac {4 \sec (e+f x) (a \sin (e+f x)+a)^{2/3}}{5 a f}-\frac {3 \sec (e+f x)}{5 f \sqrt [3]{a \sin (e+f x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^2/(a + a*Sin[e + f*x])^(1/3),x]

[Out]

(-3*Sec[e + f*x])/(5*f*(a + a*Sin[e + f*x])^(1/3)) + (11*2^(1/6)*Cos[e + f*x]*Hypergeometric2F1[1/2, 5/6, 3/2,
(1 - Sin[e + f*x])/2])/(15*f*(1 + Sin[e + f*x])^(1/6)*(a + a*Sin[e + f*x])^(1/3)) + (4*Sec[e + f*x]*(a + a*Si
n[e + f*x])^(2/3))/(5*a*f)

Rule 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[a^IntPart[n]*((a + b*Sin[c + d*x])^FracPart
[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n]), Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2791

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[b*((a + b*Sin[e +
f*x])^m/(a*f*(2*m - 1)*Cos[e + f*x])), x] - Dist[1/(a^2*(2*m - 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*((a*m -
b*(2*m - 1)*Sin[e + f*x])/Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ
[m] && LtQ[m, 0]

Rule 2934

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p +
1))), x] + Dist[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1))), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*
x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\tan ^2(e+f x)}{\sqrt [3]{a+a \sin (e+f x)}} \, dx &=-\frac {3 \sec (e+f x)}{5 f \sqrt [3]{a+a \sin (e+f x)}}+\frac {3 \int \sec ^2(e+f x) (a+a \sin (e+f x))^{2/3} \left (-\frac {a}{3}+\frac {5}{3} a \sin (e+f x)\right ) \, dx}{5 a^2}\\ &=-\frac {3 \sec (e+f x)}{5 f \sqrt [3]{a+a \sin (e+f x)}}+\frac {4 \sec (e+f x) (a+a \sin (e+f x))^{2/3}}{5 a f}-\frac {11}{15} \int \frac {1}{\sqrt [3]{a+a \sin (e+f x)}} \, dx\\ &=-\frac {3 \sec (e+f x)}{5 f \sqrt [3]{a+a \sin (e+f x)}}+\frac {4 \sec (e+f x) (a+a \sin (e+f x))^{2/3}}{5 a f}-\frac {\left (11 \sqrt [3]{1+\sin (e+f x)}\right ) \int \frac {1}{\sqrt [3]{1+\sin (e+f x)}} \, dx}{15 \sqrt [3]{a+a \sin (e+f x)}}\\ &=-\frac {3 \sec (e+f x)}{5 f \sqrt [3]{a+a \sin (e+f x)}}+\frac {11 \sqrt [6]{2} \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{15 f \sqrt [6]{1+\sin (e+f x)} \sqrt [3]{a+a \sin (e+f x)}}+\frac {4 \sec (e+f x) (a+a \sin (e+f x))^{2/3}}{5 a f}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 100, normalized size = 0.79 \begin {gather*} \frac {-22 \cos (e+f x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right )+\sqrt {2-2 \sin (e+f x)} (\sec (e+f x)+4 \tan (e+f x))}{5 f \sqrt {2-2 \sin (e+f x)} \sqrt [3]{a (1+\sin (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^2/(a + a*Sin[e + f*x])^(1/3),x]

[Out]

(-22*Cos[e + f*x]*Hypergeometric2F1[1/6, 1/2, 7/6, Sin[(2*e + Pi + 2*f*x)/4]^2] + Sqrt[2 - 2*Sin[e + f*x]]*(Se
c[e + f*x] + 4*Tan[e + f*x]))/(5*f*Sqrt[2 - 2*Sin[e + f*x]]*(a*(1 + Sin[e + f*x]))^(1/3))

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\tan ^{2}\left (f x +e \right )}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^2/(a+a*sin(f*x+e))^(1/3),x)

[Out]

int(tan(f*x+e)^2/(a+a*sin(f*x+e))^(1/3),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+a*sin(f*x+e))^(1/3),x, algorithm="maxima")

[Out]

integrate(tan(f*x + e)^2/(a*sin(f*x + e) + a)^(1/3), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+a*sin(f*x+e))^(1/3),x, algorithm="fricas")

[Out]

integral(tan(f*x + e)^2/(a*sin(f*x + e) + a)^(1/3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{2}{\left (e + f x \right )}}{\sqrt [3]{a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**2/(a+a*sin(f*x+e))**(1/3),x)

[Out]

Integral(tan(e + f*x)**2/(a*(sin(e + f*x) + 1))**(1/3), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+a*sin(f*x+e))^(1/3),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^2/(a*sin(f*x + e) + a)^(1/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^2/(a + a*sin(e + f*x))^(1/3),x)

[Out]

int(tan(e + f*x)^2/(a + a*sin(e + f*x))^(1/3), x)

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